Question

If sin(-2x0yz) = 0, use implicit differentiation to find the first partial derivatives at the point (0, 0, 0).

Answer 1

To find the first partial derivatives at the point (0, 0, 0) using implicit differentiation, differentiate the equation sin(-2x0yz) = 0 with respect to each variable, treating other variables as constants. The first partial derivatives are ∂/∂x = -2y0z*cos(-2x0yz), ∂/∂y = -2x0z*cos(-2x0yz), and ∂/∂z = -2x0y*cos(-2x0yz).

Step-by-step explanation:

To find the first partial derivatives at the point (0, 0, 0) using implicit differentiation, we can differentiate both sides of the equation sin(-2x0yz) = 0 with respect to each variable, treating the other variables as constants.

Let's start with differentiating with respect to x:

d/dx(sin(-2x0yz)) = d/dx(0)

-2y0z*cos(-2x0yz) = 0

Simplifying the equation, we have -2y0z*cos(-2x0yz) = 0.

Now, differentiate with respect to y:

d/dy(sin(-2x0yz)) = d/dy(0)

-2x0z*cos(-2x0yz) = 0

Simplifying the equation, we have -2x0z*cos(-2x0yz) = 0.

Finally, differentiate with respect to z:

d/dz(sin(-2x0yz)) = d/dz(0)

-2x0y*cos(-2x0yz) = 0

Simplifying the equation, we have -2x0y*cos(-2x0yz) = 0.

Therefore, the first partial derivatives at the point (0, 0, 0) are:

∂/∂x = -2y0z*cos(-2x0yz)

∂/∂y = -2x0z*cos(-2x0yz)

∂/∂z = -2x0y*cos(-2x0yz)